Differential equation: undetermined coeff.

1. The problem statement, all variables and given/known data
i'm stuck. could someone please help me with this? thanks!
find the general solution:

y''+2y'+y=2e^-t

2. Relevant equations

CE: r^2+2r+1=0
(r+1)^2 which means r1=r2=-1
c1e^-t+c2te^-t

3. The attempt at a solution

Y(t)=Ae^-t
Y'(t)=-Ae^-t
Y''(t)=Ae^-t
back into the equation:
Ae^-t-2Ae^-t+Ae^-t=2Ae^-t
0=Ae^-t
now im stuck! what am i doing wrong?! this is the "easiest" problem! isn't the A left there and that's how i find the GS?

Yes, typically that would be a good guess at the particular solution, but essentially what happens is that the solution space (general solution) is the superposition of the homogeneous solution and the particular solution. If you add a constant multiple of the homogeneous solution then that will get you nowhere. In other words, if your guess at the particular solution has the same form as the homogeneous solution then you will not find the other linear function in the superposition of the general solution because L1[x] + cL1[x] doesn't really mean anything.

can i have a hint? =) i just tried Ate^-t from another book that says if the assumed solution is a multiplicative function of the homogenous solution and that it should be multiplied by t^m where m is the smallest integer possible. (did i do that right?)

Alright, you were on the right track by continually adding coefficients and higher orders of t.

Did you know that you could solve an equation that only needed Ae^-t with as many higher order terms as you wanted?
Zt^n e^-t + Yt^n-1 e^-t + ... + Bt e^-t + Ae^-t would give the same thing for an equation of say y'' + y = 10e^-t as guessing the lowest order term Ae^-t.

ok great, thanks! you've been a lot of help. but i guess i still made a mistake because my guess was Ate^-t+Bt^2 e^-t+C instead of Ae^-t+Bte^-t+Ct^2e^t. so just to confirm, the guess should be At^-t + Bt^-t + Ct^2e^-t? i thought it was just to multiply the homogeneous solution by a factor of t...sorry...taking a while to digest, this stuff is still really new.

Yeah, you should have had my guess, which is really (A + Bt + Ct^2)e^-t if that helps you see it a little clearer. You just have to make sure your guess at the particular solution doesn't copy the homogeneous solution because it will not have the range (solution space) as the homogeneous.

Say you had a homogeneous solution of c1e^t + c2t^2e^t + c3t^3e^t + c4t^4e^t then your particular would have to have one higher order than the homogeneous y_p = (A + Bt +Ct^2 + Dt^3 + Et^4 + Ft^5)e^-t.